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    Mark Scheme (Results)

    June 2011

    GCE Decision D1 (6689) Paper 1

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    Edexcel is one of the leading examining and awarding bodies in the UK andthroughout the world. We provide a wide range of qualifications including

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    June 2011

    Publications Code UA027669

    All the material in this publication is copyright

    Edexcel Ltd 2011

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    EDEXCEL GCE MATHEMATICS

    General Instructions for Marking

    1. The total number of marks for the paper is 75.

    2. The Edexcel Mathematics mark schemes use the following types of marks:

    M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise

    indicated.

    A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.

    B marks are unconditional accuracy marks (independent of M marks)

    Marks should not be subdivided.

    3. Abbreviations

    These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used ifyou are using the annotation facility on ePEN.

    bod benefit of doubt

    ft follow through

    the symbol will be used for correct ft

    cao correct answer only

    cso - correct solution only. There must be no errors in this part of the question to obtain this mark

    isw ignore subsequent working

    awrt answers which round to

    SC: special case

    oe or equivalent (and appropriate)

    dep dependent

    indep independent

    dp decimal places

    sf significant figures

    The answer is printed on the paper

    The second mark is dependent on gaining the first mark

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    GCE Decision Mathematics D1 (6689) June 2011

    1

    June 2011

    Decision Mathematics D1 6689

    Mark Scheme

    QuestionNumber

    Scheme Marks

    1.

    (a) The list is not in alphabetical order. B1

    (1)

    E.g. A Quick sort

    (b) J M C B T H K R G F H M1

    C B G F H J M T K R G T

    C B F G H J M K R T B K A1

    B C F G H J K M R T F R

    B C F G H J K M R T A1Sort complete + named correctly A1= B1

    (4)

    (c) 1 10Pivot 1 = 6 Jenny reject 1 - 6

    2

    +=

    M1 A1

    7 10Pivot 2 9 Richard reject 9 - 10

    2

    + = =

    A1ft

    7 8Pivot 3 = 8 Merry reject 8

    2

    +=

    Pivot 4 = 7 Kim - name found A1(4)

    9

    Notes:

    (a) B1 CAO phonetically close

    (b) M1 Quick sort pivots, p, selected and first pass gives p.

    1A1 First two passes correct, pivots chosen consistently for third pass

    2A1 CAO Sort completed correctly

    3A1=2B1 Stop + plus correct name for their sort phonetically close

    (c) M1 Using their sorted list + choosing middle right pivots+ discarding/retaining half the

    list. It their list is not in full alphabetical order M1 only.1A1 First pass correct ie 6

    thitem for a correct list (no sticky pivots)

    2A1 Second and third passes correct ie 9th

    and 8th

    items from a correct list (no sticky

    pivots)

    3A1 CSO search complete + found

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    GCE Decision Mathematics D1 (6689) June 2011

    2

    Question

    NumberScheme Marks

    2.

    (a)(i) A tree is a connected graph with no cycles/circuit B1

    (a)(ii) A minimum spanning tree is a tree that contains all vertices and B1

    the total length of its arcs (weight of tree) is as small as possible. B1

    (3)

    (b)AB, DE, BC;

    reject AC

    BD

    reject BE, reject CE, use either EF or CFM1; A1

    A1

    (3)

    (c)

    B1 (1)

    (d) No, there are two solutions since either EF or CF should be used. B1

    (1)

    8

    Notes

    (a)1B1 Connected + no cycles

    2B1 Contains all vertices3B1 Total length of arcs used minimised or minimum weight. (Not shortest/smallest etc.)

    (b)M1 First four arcs selected correctly in correct order.

    1A1 Arcs selected correctly at correct time

    2A1 Rejections correct and at correct time

    (c)B1 CAO

    (d)B1 CAO - mark explanation must specify two arcs of 18 or two 18s or ref to EF and CF

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    GCE Decision Mathematics D1 (6689) June 2011

    3

    Question

    NumberScheme Marks

    3.

    (a) 6 5 60x y+

    2 3 12x y+ 3 2x y

    2x y

    B2,1,0

    (2)

    (b) Drawing objective line{ (0,3) (1,0)} Testing at least 2 points M1 A1

    Calculating optimal point Testing at least 3 points DM1

    ( ) 120 60917 ,3 , (7.06,3.53)17 17 17 17

    =

    A1 awrt

    (4)

    (c) 12 24024 24.7

    17 17= (awrt) B1

    (1)

    (d) (6,4) B1

    (1)

    8

    Notes: ( ) ( )3 5 24 127 7 7 73 ,1 , (3.43,1.71) 12=

    ( ) ( ) ( )10 36 10811 24 4

    13 13 13 13 13 131 ,2 , (1.85,2.77) 8.3 07692 8= =

    ( ) ( )40 204 29 3 9 34 ,6 , (4.44,6.67) 20=

    ( ) ( ) ( )9 120 60 4201 1217 17 17 17 17 177 ,3 , (7.06,3.53) 24.705882 24= =

    Notes

    (a)1B1 Any two inequalities correct, accept < and > here (but not = of course).

    2B1 All four correct. Must be and here(b)1M1 Drawing objective line or its reciprocal OR testing two vertices in the feasible region

    (see list above) points correct to 1 dp.

    1A1 Correct objective line OR two points correctly tested (1 dp ok)2DM1 Calculating optimal point either answer to 2 dp or better or using S.Es (correct 2

    equations for their point + attempt to eliminate one variable.); OR Testing three

    points correctly and optimal one to 2dp.

    2A1 CAO 2 dp or better.

    (c)B1 CAO

    (d)B1 CAO not (4,6).

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    GCE Decision Mathematics D1 (6689) June 2011

    4

    Question

    NumberScheme Marks

    4. [Given A 3 = R 4 = C 5 ]

    (a) A 1 H 2= M1 A1A 1 H 3 R 4 C 5= = = A1

    (3)

    (b) ( )A 3, C 5, H 1, J unmatched , R 4= = = = B1

    (1)

    (c) Alternating path : J 4 R 3 A 1 H 2= = = M1

    Change status : J 4 R 3 A 1 H 2= = = = A1

    A 1, C 5, H 2, J 4, R 3= = = = = A1

    (3)

    7

    (a)M1

    NotesPath from A to 2 or 5 - or vice versa

    1A1 One correct path selected OR tree showing the missing two paths only.

    2A1 Both correct paths listed separately

    (b)B1 CAO

    (c)M1 Path from J to 2 or vice versa

    1A1 Correct path including change status

    2A1 CAO must follow through from stated path.

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    GCE Decision Mathematics D1 (6689) June 2011

    5

    Question

    NumberScheme Marks

    5.

    (a) AC DF 9 13 22+ = + = M1 A1AD CF 16 8 24+ = + = A1AF CD 17 7 24+ = + = A1Repeat arcs AC, DG and GF A1ft

    (5)

    (b) E.g. ADCACGDGFGECBEFBA B1

    ( )Length of route 98 22 120 km= + = B1ft

    (2)

    (c) CF (8) is the shortest link between 2 odd nodes excluding D

    Repeat CF (8) since this is the shortest path excluding D. M1

    We finish at A A1ftLength of route = 98 + 8 = 106 (km) A1ft

    (3)

    10

    Notes

    (a)M1 Three pairings of their four odd nodes

    1A1 One row correct including pairing and total

    2A1 Two rows correct including pairing and total

    3A1 Three rows correct including pairing and total

    4A1ft Their smallest repeated arcs stated accept DGF or arcs clear from selected row.

    (b)1B1 Correct route any start point, 17 nodes, AC, DG and GF repeated

    2B1ft CAO 98 + their least out of a choice of at least 2.

    (c)M1 Attempting just one repeated path excluding D; accept AC, AF and CF listed

    1A1ft A and their least repeat [should be CF (CEF)] clearly stating this as least

    2A1ft 98 + their least from their working in (a)

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    GCE Decision Mathematics D1 (6689) June 2011

    6

    Question

    NumberScheme Marks

    6.

    (a) M1

    A1

    (ABCD)

    A1ft

    (EF)

    A1ft

    (GH)

    ACDFEGH A1

    Length 71 (km) A1ft

    (6)

    (b) E.g. 71 12 = 59 GH 49 10 = 39 FE 24 13 = 11 CD

    59 10 = 49 EG 39 15 = 24 DF 11 11 = 0 AC

    Or Trace back from H including arc XY if (Y already lies on the path and) the

    difference of the final values of X and Y equals weight of arc XY.

    B2,1,0

    (2)

    (c) ACBEGH B1

    Length 72 (km) B1

    (2)

    10

    Notes

    (a)M1 Big replaced by smaller at least once at B or D or E or G or H

    1A1 A, B, C, D boxes all correct, condone lack of 0 in s working value

    2A1ft E and F ft correctly

    3A1ft G and H ft correctly

    4A1 CAO

    5A1ft ft on their final value.(b)1B1 Attempting an explanation, at least 3 stages or one half of general explanation

    2B1 Correct explanation all six stages, both halves of explanation

    (c)1B1 CAO

    2B1 CAO

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    GCE Decision Mathematics D1 (6689) June 2011

    7

    Question

    Number

    Scheme Marks

    7.

    (a) Activity Proceeded by Activity Proceeded by Activity Proceeded by

    (A) (-) E A B I C D E(B) (-) (F) (B) J C D E

    C A B (G) (B) K F H I

    (D) (B) H C D L F G H I B3,2,1,0

    (3)

    (b)

    M1 A1

    M1 A1

    (4)

    (c) Critical activities are B D J H L M1 A1(2)

    (d)

    M1 A1

    M1 A1

    (4)

    E

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    GCE Decision Mathematics D1 (6689) June 2011

    8

    Question

    Number

    SchemeMarks

    (e) E.g.

    Between time 7 and 16, 3 workers could do 3 x 9 = 27 days work.

    Activities C, D, E, F, G, H, I and 4 days of J need to be doneThis totals 31 days work.

    So it is not possible to complete the project with three workers.

    B3,2,1,0

    OR (3)

    If three workers are used three activities H, J and I need to happen at time 13.5, this

    reduces the float on F and G, meaning that at 10.5 D, C, F and G need to be

    happening. Our initial assumption is incorrect hence four workers are needed.

    16

    (a)1B1

    NotesAny two rows correct

    2B1 Any four rows correct3B1 All seven rows correct

    (b)1M1 All top boxes complete, values generally increasing left to right, condone one rogue

    1A1 CAO

    2M1 All bottom boxes complete, values generally decreasing R to L, condone one rogue

    2A1 CAO

    (c)M1 Accept dummies, repeats and condone one absence or one extra; or BDHL or BDJ

    A1 CAO (dummies and repeats ok)

    (d)1M1 At least 9 activities including at least 4 floats. Do not accept scheduling diagram.

    1A1 Critical activities dealt with correctly

    2M1All 12 activities including at least 7 floats2A1 Non-critical activities dealt with correctly.

    (e)1B1 Attempt at explanation one correct idea.

    2B1 Good explanation, some imprecise or vague statements give bod

    3B1 Fully correct explanation. No bod needed

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    GCE Decision Mathematics D1 (6689) June 2011

    9

    Question

    Number

    Scheme Marks

    8. Letx be the number of type A radios and y be the number of type B radios. B1

    ( )Maximise P 15 12x y= + B1Subject to

    50x B1

    [ ]15 ( ) (accept ) 4 x y x y x+ < < B1

    [ ]25 ( ) (accept ) 2 3 x y x y x+ > > B1

    3 2 200x y+ B1

    0y B1

    7

    Notes1B1 Definingx andy; Must see number of

    2B1 CAO objective function 15 12x y+

    3B1 CAO 50x 4B1 CAO o.e 1

    5( ) 4 x y x y x+ < <

    5B1 CAO o.e 23( ) 2 3 x y x y x+ > >

    6B1 CAO o.e 3 2 200x y+

    7B1 CAO 0y

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    GCE Decision Mathematics D1 (6689) June 2011

    10

    Question

    Number

    Scheme Marks

    Q1 b Additional solutions

    Quick sort middle left

    J M C B T H K R G F T M1

    J M C B H K R G F T H

    C B G F H J M K R T B M A1

    B C G F H J K M R T G J

    B C F G H J K M R T C

    B C F G H J K M R T A1

    Quick sort complete A1

    Bubble sort left to right

    J M C B T H K R G F

    J C B M H K R G F T T in place, consistent direction M1

    C B J H K M G F R T

    B C H J K G F M R T Passes 1 and 2 correct A1

    B C H J G F K M R T

    B C H G F J K M R T

    B C G F H J K M R T

    B C F G H J K M R T Sort correct A1

    Bubble sort complete sort named correctly + stop A1

    Bubble right to left

    J M C B T H K R G F

    B J M C F T H K R G B in place, consistent direction M1

    B C J M F G T H K R

    B C F J M G H T K R Passes 1 and 2 correct A1

    B C F G J M H K T R

    B C F G H J M K R T

    B C F G H J K M R T Sort correct A1

    Bubble sort complete sort named correctly + stop A1

    Sorting into reverse alphabetical order this is acceptable

    e.g. Quick sort middle right

    J M C B T H K R G F H M1

    J M T K R H C B G F T G

    T J M K R H G C B F K B A1

    T M R K J H G C F B R F

    T R M K J H G F C B A1

    Quick sort complete A1

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